Answer

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**Hint:**Since $|t| = 1$ we will try to separate the factor of ‘t’ from the expression $z = \dfrac{{at + b}}{{t - c}}$.

After separating the term ‘t’ we will take the modulus function of both the side and will eliminate ‘t’ using$ |t| = 1$.

**Complete step by step solution:**

Given data:

$

a,\;b,\;c,\; \text{and}\:{\text{ t}}\:{\text{are }}\:{\text{complex }}\:{\text{numbers }} \\

|t| = 1 \\

|t| \ne |c| \\

z = \dfrac{{at + b}}{{t - c}},z = x + iy. \\

$

Now solving the given equation, i.e.

$

z = \dfrac{{at + b}}{{t - c}} \\

\Rightarrow z(t - c) = at + b \\

\Rightarrow zt - at = b + cz \\

\Rightarrow t(z - a) = b + cz \\

\Rightarrow t = \dfrac{{b + cz}}{{z - a}} \\

$

Now, applying the modulus function on both sides of the equation, we obtain

$

\left| t \right| = \left| {\dfrac{{b + cz}}{{z - a}}} \right| \\

\sin ce{\text{ }}\left| t \right| = 1 \\

1 = \left| {\dfrac{{b + cz}}{{z - a}}} \right| \\

$

${\text{Now have also condition on}}$ ${\text{'c'}}$that \[\left| {\text{c}} \right| \ne {\text{1}}\]

$

1 = \left| {\dfrac{{c\left( {\dfrac{b}{c} + z} \right)}}{{z - a}}} \right| \\

we{\text{ }}know{\text{ }}that,{\text{ }}\left| {ab} \right| = \left| a \right|\left| b \right| \\

\Rightarrow 1 = \left| {\dfrac{{\dfrac{b}{c} + z}}{{z - a}}} \right|\left| c \right| \\

\Rightarrow \dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + \dfrac{b}{c}}}{{z - a}}} \right| \\

$

Since b, c are complex numbers, then $\dfrac{b}{c}$ be also a complex number let ‘p’

$\dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + p}}{{z - a}}} \right|$

We know that,

$\left| {\dfrac{{z - \alpha }}{{z - \beta }}} \right| = A$

When $A \ne 1$, then the locus of z is a circle.

Here also $\left| c \right| \ne 1$ i.e.,

$\left| {\dfrac{{z + p}}{{z - a}}} \right| \ne 1$

Therefore the locus of z is a circle.

**(C) The circle is the correct option.**

**Note:**It is well known that $\left| z \right| = a,$ ‘a’ being a constant is an equation of a circle.

From the equation

$z = \dfrac{{at + b}}{{t - c}}$

Applying the modulus function on both sides of the equation, we obtain

$\left| z \right| = \left| {\dfrac{{at + b}}{{t - c}}} \right|$

We can say that the right-hand side will come to be a constant as a, b, c and t are some fixed complex number.

We can write it as

$\left| z \right| = f$, where f is constant, comes out to be an equation of a circle.

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